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aroub Group Title

Why do we have to put an absolute value in the final answer?!

  • 2 years ago
  • 2 years ago

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  1. estudier Group Title
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    final answer to what?

    • 2 years ago
  2. aroub Group Title
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    \[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}\]

    • 2 years ago
  3. estudier Group Title
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    Can't even see that, is it 10 th root?

    • 2 years ago
  4. aroub Group Title
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    Yup

    • 2 years ago
  5. akash_809 Group Title
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    \[\sqrt[10]{x^10}\]

    • 2 years ago
  6. akash_809 Group Title
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    see moderator has warned me now...whats the way to delet ur account...can somebody give the link...i m going frm here

    • 2 years ago
  7. aroub Group Title
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    What? O.o

    • 2 years ago
  8. akash_809 Group Title
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    can u pls give me the link...i want to delet my account..i can't find it anywhere

    • 2 years ago
  9. aroub Group Title
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    I don't know.. You have to contact a moderator

    • 2 years ago
  10. akash_809 Group Title
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    cmon moderator show up now...

    • 2 years ago
  11. phi Group Title
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    The only reason I can think of is they want to make sure you get a real and postiive (as opposed to complex) root. There will be 10 different roots.

    • 2 years ago
  12. TuringTest Group Title
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    @akash_809 you cannot delete OS accounts

    • 2 years ago
  13. aroub Group Title
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    But you can write it without absolute value, right? Because (or this is what I've been taught) there's x^3 in the radicand so it's definitely positive..

    • 2 years ago
  14. TuringTest Group Title
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    the simple reason is that\[x^{10}=(-x)^{10}\]so \[\large\sqrt[10]{x^{10}}\]will give you the same answer whether x is negative or positive

    • 2 years ago
  15. akash_809 Group Title
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    what...thats cynical...open study community is surely going nowher with all these warnings and suspensions...and to top that off , you even cant delete your account...what is this open study team please response...i wnat to know how can i delete my account

    • 2 years ago
  16. TuringTest Group Title
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    absolute value of x is actually defined in 2 ways: 1) distance from the point x=0 on the number line 2) \(|x|=\sqrt{x^2}\)

    • 2 years ago
  17. aroub Group Title
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    Thanks :))

    • 2 years ago
  18. TuringTest Group Title
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    so since\[\sqrt{(-x)^2}=\sqrt{x^2}=|x|\]then\[\sqrt[10]{x^{10}}=\sqrt[10]{(-x)^{10}}=|x|\](because x to any even power is positive, whether x>0 or x<0)

    • 2 years ago
  19. TuringTest Group Title
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    does that help you @aroub ?

    • 2 years ago
  20. TuringTest Group Title
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    welcome :D

    • 2 years ago
  21. aroub Group Title
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    In this case I totally understand why. But in the question itself there's an odd exponent in the radicand so it's definitely positive right? So then no need for an absolute value..

    • 2 years ago
  22. akash_809 Group Title
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    @TuringTest root of negative number is not defined...it will result in iota value...so -x wont be the answer i think

    • 2 years ago
  23. TuringTest Group Title
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    not necessarily, let's assume x<0...

    • 2 years ago
  24. aroub Group Title
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    If you assumed that, the question will be wrong.

    • 2 years ago
  25. aroub Group Title
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    -I think-

    • 2 years ago
  26. akash_809 Group Title
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    yes it is...\[lets take an example......\sqrt{-2} is \not defined unless you are taking imaginary numbers into account then u wl get +-1.414 i\]

    • 2 years ago
  27. TuringTest Group Title
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    the root of a negative number is not undefined, it is imaginary...\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}\]

    • 2 years ago
  28. akash_809 Group Title
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    sry reply got messed up i was just saying that square root of -2 is not defined unless you consider imaginary numbers as wel..

    • 2 years ago
  29. aroub Group Title
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    Put imaginary numbers on a side for now..

    • 2 years ago
  30. TuringTest Group Title
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    imaginary numbers are defined... now if x<0, then x^3<0 that means that (x^3)(x^3)>0 which is what we would expect given that x^6>0 since all numbers to a positive exponent are >0

    • 2 years ago
  31. TuringTest Group Title
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    another way to see it, let's assume x>0\[\sqrt[10]{x^4} \times \sqrt[10]{x^3} \times \sqrt[10]{x^3}=\sqrt[10]{x^4}\times\sqrt[10]{x^3\cdot x^3}=\sqrt[10]{x^{10}}=|x|\]now we do it with the negative of x\[\sqrt[10]{(-x)^4} \times \sqrt[10]{(-x)^3} \times \sqrt[10]{(-x)^3}=\sqrt[10]{(-x)^4}\times\sqrt[10]{(-x)^3\cdot (-x)^3}\]\[=\sqrt[10]{(-x)^{10}}=\sqrt[10]{x^{10}}=|x|\]so you see, after all is said and done we do not know whether the x we had at the beginning was positive or negative, hence we can only conclude that the answer is |x|

    • 2 years ago
  32. aroub Group Title
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    Okay take this as another example: \[4\sqrt[4]{5a^2}\times 8\sqrt[4]{125a}\times 2\sqrt[4]{a}\] Here, a does not need an absolute value. Why? You can assume that a>0

    • 2 years ago
  33. akash_809 Group Title
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    yes you would have to assume a >0

    • 2 years ago
  34. TuringTest Group Title
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    let's just look at powers of a...

    • 2 years ago
  35. TuringTest Group Title
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    if a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\](since we already knew a>0 now with the negative of a...

    • 2 years ago
  36. akash_809 Group Title
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    yea i m waiting for that negative values of a...what would it give

    • 2 years ago
  37. aroub Group Title
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    Aoh, so you have to try the negative and the positive?

    • 2 years ago
  38. TuringTest Group Title
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    \[\sqrt[4]{(-a)^2}\times\sqrt[4]{-a}\times\sqrt[4]{-a}=\sqrt[4]{a^4}=|a|\neq a\]because |a|>0, but we just assumed a<0 ! a contradiction, hence the assumption makes a difference

    • 2 years ago
  39. TuringTest Group Title
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    sorry, I mean we took the negative of a...

    • 2 years ago
  40. TuringTest Group Title
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    so \[|a|\neq a\]

    • 2 years ago
  41. TuringTest Group Title
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    let me write that again, gotta be careful here

    • 2 years ago
  42. akash_809 Group Title
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    that's the problem...the thing you did here is called fallacy of maths...http://en.wikipedia.org/wiki/Mathematical_fallacy ....search invalid roots in this article

    • 2 years ago
  43. TuringTest Group Title
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    assume a>0\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|=a\]assume a<0, then\[\sqrt[4]{a^2}\times\sqrt[4]a\times\sqrt[4]a=\sqrt[4]{a^4}=|a|\neq a\]since we assumed a<0

    • 2 years ago
  44. TuringTest Group Title
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    @akash_809 yes but at least one of these numbers is positive because it is raised to an even power, so I have committed no fallacy

    • 2 years ago
  45. akash_809 Group Title
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    did u went through the link......the proofs given like \[1=\sqrt{1}=\sqrt{(-1)(-1)}=\sqrt{-1}\sqrt{-1}=i.i=-1 ....so u hv 1=-1...and certainly \it is \not valid..\]

    • 2 years ago
  46. TuringTest Group Title
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    same thing with the earlier problem...

    • 2 years ago
  47. akash_809 Group Title
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    and 1=-1....surely not right...

    • 2 years ago
  48. TuringTest Group Title
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    yes, but\[\sqrt{(-1)^2}\cdot\sqrt{-1}=\sqrt1\cdot\sqrt{-1}\] is not the same as\[\sqrt{-1}\cdot\sqrt{-1}\]

    • 2 years ago
  49. TuringTest Group Title
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    notice we have a^2 in the above, so we know *that* part is positive

    • 2 years ago
  50. TuringTest Group Title
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    this turned out to be more subtle than I bet you realized @aroub :)

    • 2 years ago
  51. aroub Group Title
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    lol, yeah true xD ALWAYS thought absolute values are so easy! And whenever someone asks me about absolute values I answer until this question came up -.-' This question is a bit confusing >.< Well, THANKS LOOAADSSSSSS :D

    • 2 years ago
  52. TuringTest Group Title
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    welcome :)

    • 2 years ago
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